Density argument in the definition of weak solution of PDEs
$begingroup$
I don't remember how to properly prove the following.
Let us consider the following PDE (transport equation):
$$
(1) quad left{begin{array}{l}
u_t+u_x=0, \
u(t,0)=h(t), \
u(0,x)=u_0(x).
end{array}right.
$$
where $t in (0,T)$ and $x in (0,1)$, and $h in L^2_{mathrm{loc}}(0,+infty)$ and $u_0 in L^2(0,1)$.
Then, the classical way to define the notion of weak solution is to multiply (1) by $varphi in C^1([0,T]times[0,1])$ and do integrations by parts.
This leads to the following definition: $u$ is a solution to (1) if $u in C^0([0,+infty);L^2(0,1))$ and satisfies, for every $T>0$,
$$
(2) quad
int_0^1 u(T,x)varphi(T,x) , dx - int_0^1 u_0(x)varphi(0,x) , dx
+int_0^T int_0^1 u(t,x) left(-varphi_t(t,x)-varphi_x(t,x)right), dx dt
-int_0^T h(t) varphi(t,0) , dt=0,
$$
for every $varphi in D$, where
$$D=left{varphi in C^1([0,T]times[0,1]), quad varphi(t,1)=0, quad forall t in [0,T]right}.$$
Now, we see that each term makes sense in (2) if $varphi in E$, where
$$E=left{varphi in H^1(0,T;L^2(0,1)) cap L^2(0,T;H^1(0,1)), quad varphi(t,1)=0 mbox{ a.e. } t in (0,T)right}.$$
My question is: How to prove that a function $u in C^0([0,+infty);L^2(0,1))$ is a solution to (1) if, and only if, it satisfies (2) for every $varphi in E$ (and every $T>0$) ?
pde
asked Dec 29 '18 at 13:08
perturbationperturbation
828
$endgroup$
add a comment |
$begingroup$
I don't remember how to properly prove the following.
Let us consider the following PDE (transport equation):
$$
(1) quad left{begin{array}{l}
u_t+u_x=0, \
u(t,0)=h(t), \
u(0,x)=u_0(x).
end{array}right.
$$
where $t in (0,T)$ and $x in (0,1)$, and $h in L^2_{mathrm{loc}}(0,+infty)$ and $u_0 in L^2(0,1)$.
Then, the classical way to define the notion of weak solution is to multiply (1) by $varphi in C^1([0,T]times[0,1])$ and do integrations by parts.
This leads to the following definition: $u$ is a solution to (1) if $u in C^0([0,+infty);L^2(0,1))$ and satisfies, for every $T>0$,
$$
(2) quad
int_0^1 u(T,x)varphi(T,x) , dx - int_0^1 u_0(x)varphi(0,x) , dx
+int_0^T int_0^1 u(t,x) left(-varphi_t(t,x)-varphi_x(t,x)right), dx dt
-int_0^T h(t) varphi(t,0) , dt=0,
$$
for every $varphi in D$, where
$$D=left{varphi in C^1([0,T]times[0,1]), quad varphi(t,1)=0, quad forall t in [0,T]right}.$$
Now, we see that each term makes sense in (2) if $varphi in E$, where
$$E=left{varphi in H^1(0,T;L^2(0,1)) cap L^2(0,T;H^1(0,1)), quad varphi(t,1)=0 mbox{ a.e. } t in (0,T)right}.$$
My question is: How to prove that a function $u in C^0([0,+infty);L^2(0,1))$ is a solution to (1) if, and only if, it satisfies (2) for every $varphi in E$ (and every $T>0$) ?
pde
asked Dec 29 '18 at 13:08
perturbationperturbation
828
$endgroup$
add a comment |
$begingroup$
I don't remember how to properly prove the following.
Let us consider the following PDE (transport equation):
$$
(1) quad left{begin{array}{l}
u_t+u_x=0, \
u(t,0)=h(t), \
u(0,x)=u_0(x).
end{array}right.
$$
where $t in (0,T)$ and $x in (0,1)$, and $h in L^2_{mathrm{loc}}(0,+infty)$ and $u_0 in L^2(0,1)$.
Then, the classical way to define the notion of weak solution is to multiply (1) by $varphi in C^1([0,T]times[0,1])$ and do integrations by parts.
This leads to the following definition: $u$ is a solution to (1) if $u in C^0([0,+infty);L^2(0,1))$ and satisfies, for every $T>0$,
$$
(2) quad
int_0^1 u(T,x)varphi(T,x) , dx - int_0^1 u_0(x)varphi(0,x) , dx
+int_0^T int_0^1 u(t,x) left(-varphi_t(t,x)-varphi_x(t,x)right), dx dt
-int_0^T h(t) varphi(t,0) , dt=0,
$$
for every $varphi in D$, where
$$D=left{varphi in C^1([0,T]times[0,1]), quad varphi(t,1)=0, quad forall t in [0,T]right}.$$
Now, we see that each term makes sense in (2) if $varphi in E$, where
$$E=left{varphi in H^1(0,T;L^2(0,1)) cap L^2(0,T;H^1(0,1)), quad varphi(t,1)=0 mbox{ a.e. } t in (0,T)right}.$$
My question is: How to prove that a function $u in C^0([0,+infty);L^2(0,1))$ is a solution to (1) if, and only if, it satisfies (2) for every $varphi in E$ (and every $T>0$) ?
pde
asked Dec 29 '18 at 13:08
perturbationperturbation
828
$endgroup$
I don't remember how to properly prove the following.
Let us consider the following PDE (transport equation):
$$
(1) quad left{begin{array}{l}
u_t+u_x=0, \
u(t,0)=h(t), \
u(0,x)=u_0(x).
end{array}right.
$$
where $t in (0,T)$ and $x in (0,1)$, and $h in L^2_{mathrm{loc}}(0,+infty)$ and $u_0 in L^2(0,1)$.
Then, the classical way to define the notion of weak solution is to multiply (1) by $varphi in C^1([0,T]times[0,1])$ and do integrations by parts.
This leads to the following definition: $u$ is a solution to (1) if $u in C^0([0,+infty);L^2(0,1))$ and satisfies, for every $T>0$,
$$
(2) quad
int_0^1 u(T,x)varphi(T,x) , dx - int_0^1 u_0(x)varphi(0,x) , dx
+int_0^T int_0^1 u(t,x) left(-varphi_t(t,x)-varphi_x(t,x)right), dx dt
-int_0^T h(t) varphi(t,0) , dt=0,
$$
for every $varphi in D$, where
$$D=left{varphi in C^1([0,T]times[0,1]), quad varphi(t,1)=0, quad forall t in [0,T]right}.$$
Now, we see that each term makes sense in (2) if $varphi in E$, where
$$E=left{varphi in H^1(0,T;L^2(0,1)) cap L^2(0,T;H^1(0,1)), quad varphi(t,1)=0 mbox{ a.e. } t in (0,T)right}.$$
My question is: How to prove that a function $u in C^0([0,+infty);L^2(0,1))$ is a solution to (1) if, and only if, it satisfies (2) for every $varphi in E$ (and every $T>0$) ?
pde
pde
asked Dec 29 '18 at 13:08
perturbationperturbation
828
asked Dec 29 '18 at 13:08
perturbationperturbation
828
asked Dec 29 '18 at 13:08
perturbationperturbation
828
asked Dec 29 '18 at 13:08
perturbationperturbation
828
asked Dec 29 '18 at 13:08
perturbationperturbation
828
828
add a comment |
add a comment |
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